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Capacitor discharge & charging, graphs. watch

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    Hi guys


    I'm looking at the graph for current whilst charging, a capacitor much like:



    Top = discharging
    Bottom = charging

    [Obviously its current not voltage we're dealing with].

    This doesn't make sense, if current is charge/time or the rate of charge flowing, then why is at t=0 whilst the capacitor is being charged at a maximum rate before electrostatic repulsion comes into play, a minimum on the graph? Then it starts to increase? Surely it should already be at a maximum and decreasing?

    Vice versa with discharging, I can't understand why current is maxiumum when there is no charge flowing around the circuit as both potential differences across the battery and capacitor are the same?

    anyone lend us a hand?
    cheers
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    (Original post by TSB01)
    [Obviously its current not voltage we're dealing with].
    And that's why it appears not to make sense!
    What is V the voltage across in these graphs?

    The capacitor will in both cases be part of a circuit with resistance in it, and, in the case of charging, also a battery of some sort, with its own pd.
    You have to look at the whole circuit.

    When discharging, the pd on the capacitor drives the current through the external circuit. When pd is a max on the capacitor, current is max.
    Your first graph shows pd on capacitor with time. The current would follow a similar pattern.

    When charging, the current is driven by the battery you connect to the capacitor. Initially there is no pd on the capacitor (your graph shows this) and there is no pd opposing the charging current. The current is a max at the start. As the capacitor charges, the pd on it opposes the flow of charge and the current falls.
    Your graph shows the pd increasing on the capacitor. The current would look like the top graph.
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    (Original post by Stonebridge)
    And that's why it appears not to make sense!
    What is V the voltage across in these graphs?

    The capacitor will in both cases be part of a circuit with resistance in it, and, in the case of charging, also a battery of some sort, with its own pd.
    You have to look at the whole circuit.

    When discharging, the pd on the capacitor drives the current through the external circuit. When pd is a max on the capacitor, current is max.
    Your first graph shows pd on capacitor with time. The current would follow a similar pattern.

    When charging, the current is driven by the battery you connect to the capacitor. Initially there is no pd on the capacitor (your graph shows this) and there is no pd opposing the charging current. The current is a max at the start. As the capacitor charges, the pd on it opposes the flow of charge and the current falls.
    Your graph shows the pd increasing on the capacitor. The current would look like the top graph.
    Alright, so I get the first bit you describe - when P.D. is maximum on the capacitor in the secondary circuit linked with some kind of resistor, then it drives a current through into the resistor (not back round the circuit because of the dielectric) causing it to dissipate the PD or energy stored previously.

    When charging however I can't get it, the PD on the capacitor is zero, so current is max without any repulsion from the PD on the capacitor, and this begins to shift until the capacitor has the same PD as the battery, and no charge is flowing - but this contradicts the discharge because it starts at a maximum with PD?. And my textbook clearly shows current and PD following an exact trend when charging as well as discharging with each other, surely when charging: current must have a e^-x relationship must like discharging?
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    In both cases the current is a max to start with and falls exponentially.
    The graphs you give are for voltage on the capacitor, not current.
    When discharging the voltage on the capacitor falls exponentially.
    When charging the voltage on the capacitor rises as shown in your second graph.
    The current when discharging follows V. (It's the only V)
    The current when charging is determined by E - V where V is the pd on the capacitor and E the emf of the cell. The pd on the capacitor is opposing the charging current. As V increases this term decreases, and so does the current.
    I'm not sure what your book is saying. Check what voltage and what current it is actually referring to. It should tell you the same as I just did.
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    (Original post by Stonebridge)
    In both cases the current is a max to start with and falls exponentially.
    The graphs you give are for voltage on the capacitor, not current.
    When discharging the voltage on the capacitor falls exponentially.
    When charging the voltage on the capacitor rises as shown in your second graph.
    The current when discharging follows V. (It's the only V)
    The current when charging is determined by E - V where V is the pd on the capacitor and E the emf of the cell. The pd on the capacitor is opposing the charging current. As V increases this term decreases, and so does the current.
    I'm not sure what your book is saying. Check what voltage and what current it is actually referring to. It should tell you the same as I just did.
    This is exactly what I'm thinking, my CGP book does indeed look like its telling some lies.

    It has a graph which is showing two graphs "Charging" and "Discharging", clearly showing t on the x axis, and "I or V" on the y (implying that both charging and discharging graphs for current and voltage are exactly the same). **** sake.
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    It sounds very odd to me.
    I, the charging current, and V, the pd on the capacitor, definitely do not follow the same curve. As the capacitor voltage rises, the charging current falls.
    Have you asked your teacher about this? What does he/she say.
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    (Original post by TSB01)
    This is exactly what I'm thinking, my CGP book does indeed look like its telling some lies.

    It has a graph which is showing two graphs "Charging" and "Discharging", clearly showing t on the x axis, and "I or V" on the y (implying that both charging and discharging graphs for current and voltage are exactly the same). **** sake.

    Have you got the CGP free sample copy? Lol, I borrowed it from the school, and it has quite a few mistakes (I asked the teacher, since I was as confused as you are :P).

    It says that for one graph, it can either be I or V on the Y axis, it is wrong. Only V can be on the Y axis.

    And another is with Magnetic Flux (which is also wrong in the book) if you want to know...
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    (Original post by 2710)
    Have you got the CGP free sample copy? Lol, I borrowed it from the school, and it has quite a few mistakes (I asked the teacher, since I was as confused as you are :P).

    It says that for one graph, it can either be I or V on the Y axis, it is wrong. Only V can be on the Y axis.

    And another is with Magnetic Flux (which is also wrong in the book) if you want to know...
    Nah, its the A2 Level Physics - Exam Board AQA A "The Revision Guide". Dark purple book with the mysterious looking fellow in the top right. I'll have to check about that Magnetic Flux thing later on aswell, can't trust this book anymore :p: .


    (Original post by Stonebridge)
    It sounds very odd to me.
    I, the charging current, and V, the pd on the capacitor, definitely do not follow the same curve. As the capacitor voltage rises, the charging current falls.
    Have you asked your teacher about this? What does he/she say.
    Yeah, when I was writing revision notes I checked my graphs and was like.. "what? Hold on..". I'll check with him tommorrow and post back. :rolleyes:
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    (Original post by TSB01)
    This is exactly what I'm thinking, my CGP book does indeed look like its telling some lies.

    It has a graph which is showing two graphs "Charging" and "Discharging", clearly showing t on the x axis, and "I or V" on the y (implying that both charging and discharging graphs for current and voltage are exactly the same). **** sake.
    A possible explanation is a typo, and it should read "Q or V".
    The charge on a capacitor is proportional to the pd. Q=CV.
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    (Original post by TSB01)
    Nah, its the A2 Level Physics - Exam Board AQA A "The Revision Guide". Dark purple book with the mysterious looking fellow in the top right. I'll have to check about that Magnetic Flux thing later on aswell, can't trust this book anymore :p: .




    Yeah, when I was writing revision notes I checked my graphs and was like.. "what? Hold on..". I'll check with him tommorrow and post back. :rolleyes:
    Yeh, thats the one I got, but my one is a free sample copy, so its kinda cut down a bit.

    I already asked my teacher on that. It is wrong. Basically, that graph you pointed out in the book, that says I or V on the Y axis, that graph is right for the V, but if you want the I, just flip the graph upside down and it will be correct.

    The other one that is incorrect is Magnetic spinny thingy.

    Look for a little diagram that has a red wave at the top and a blue on at the bottom. And in between them, it will have a little diagram of magnets twisting. The twistings are all 90 degrees out.

    Ie, when the magnet cuts parallel to the field, the emf should be a max and it should be least emf when it is perpendicular. Check before you take my word for it.
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    (Original post by 2710)
    Yeh, thats the one I got, but my one is a free sample copy, so its kinda cut down a bit.

    I already asked my teacher on that. It is wrong. Basically, that graph you pointed out in the book, that says I or V on the Y axis, that graph is right for the V, but if you want the I, just flip the graph upside down and it will be correct.

    The other one that is incorrect is Magnetic spinny thingy.

    Look for a little diagram that has a red wave at the top and a blue on at the bottom. And in between them, it will have a little diagram of magnets twisting. The twistings are all 90 degrees out.

    Ie, when the magnet cuts parallel to the field, the emf should be a max and it should be least emf when it is perpendicular. Check before you take my word for it.
    Gotcha, yeah I also saw my physics teacher today comfirming our suspicions :p: .

    I'll check it out, cheers again.
 
 
 
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